A rope is wrapped several times around a cylinder of radius 0.2 m and mass 30 kg. What is the angular acceleration of the cylinder if the tension in the rope is 40 N and it turns without friction? (13.3 rad s⁻²)

Given:

Radius of the cylinder = r = 0.2 m
Mass of the cylinder = m = 30 kg
Tension in the rope = T = F= 40 N

To Find:

Angular acceleration = 𝛂 = ?

Solution:

The formula for Angular acceleration in terms of the moment of inertia and torque (τ) is

𝛂 = `\frac {τ}{I}` --------------(1)

So we have to find first the torque τ and moment of inertia I from the given data

Torque τ = r F = 0.2 m ｘ 40 N = 8 Nm

Moment of inertia of the cylinder

I = `\frac {1}{2}`mr² = `\frac {1}{2}` ｘ 30 kg ｘ (0.2 m)² 

I = 15 kg ｘ 0.04 m² = 0.6 kg m²

By putting values in equation (1) we have

𝛂 = `\frac {τ}{I}`

𝛂 = `\frac {8 Nm}{0.6 kg m²}`

𝛂 = 13.3 rad s⁻¹ --------------Ans.



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